SOLUTION: Please help me with this equation: {{{ 2(cos(x))^2=1-sin(x) }}} So far I've gotten down to {{{ 2-2(sin(x))^2+sin(x)-1=0 }}} but I don't know what to do from here.

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me with this equation: {{{ 2(cos(x))^2=1-sin(x) }}} So far I've gotten down to {{{ 2-2(sin(x))^2+sin(x)-1=0 }}} but I don't know what to do from here.      Log On


   

Question 1027244: Please help me with this equation: +2%28cos%28x%29%29%5E2=1-sin%28x%29+
So far I've gotten down to +2-2%28sin%28x%29%29%5E2%2Bsin%28x%29-1=0+ but I don't know what to do from here.
Answer by ikleyn(53763) About Me  (Show Source):
You can put this solution on YOUR website!
.
Please help me with this equation: +2%28cos%28x%29%29%5E2=1-sin%28x%29+
So far I've gotten down to +2-2%28sin%28x%29%29%5E2%2Bsin%28x%29-1=0+ but I don't know what to do from here.
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+2%28cos%28x%29%29%5E2 = 1-sin%28x%29+,

2-2%2Asin%5E2%28x%29+%2B+sin%28x%29-1 = 0,       (It is what your got, and it is correct !)

Next introduce new variable t = sin(x) for shortness. You will get

2+-+2t%5E2+%2B+t+-+1 = 0,

2t%5E2+-+t+-+1 = 0.

Solve this quadratic by applying the quadratic fotmula 

t%5B1%2C2%5D = %281+%2B-+sqrt%281+%2B+4%2A2%2A1%29%29%2F4 = %281+%2B-+3%29%2F4.


One root is  t%5B1%5D = 1  --->  sin(x) = 1  --->  x = pi%2F2+%2B+2kpi,  k = 0, +/-1, +/-2, . . . 


The other root is  t%5B2%5D = -1%2F2  --->  sin(x) = -1%2F2  --->  x = 7pi%2F6+%2B+2kpi,  k = 0, +/-1, +/-2, . . .